Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> pii;
int INF = 1000000000;
vector<int> dijkstra(int s, int v, vector< vector< pair<int,int> > > G){
  //s: テ・ツァツ凝ァツつケ, v: テゥツ?づァツつケテヲツ閉ー, G: テゥツ堋」テヲツ篠・テ」ツδェテ」ツつケテ」ツδ暗」ツ??」ツつウテ」ツつケテ」ツδ暗」ツ?ッテヲツュツ」テ」ツ?ァテ・ツ?・テ」ツつ古」ツ?ヲテ」ツ?甘」ツ??
  priority_queue< pair<int, int> > wait;
  vector<int> result(v, INF);
  result[s] = 0;
  wait.push(make_pair(0, s));
  while(!wait.empty()){
    int nowpoint = wait.top().second;
    int nowcost = wait.top().first;
    wait.pop();
    if(nowcost <= result[nowpoint]){
      for(int i = 0; i < G[nowpoint].size(); ++i){
	int nextpoint = G[nowpoint][i].second;
	int nextcost = -nowcost + G[nowpoint][i].first;
	if(nextcost < result[nextpoint]){
	  wait.push(make_pair(-nextcost, nextpoint));
	  result[nextpoint] = nextcost;
	}
      }
    }
  }
  return result;
}

long long int count(vector< vector<pii> > &G, int s, int t, int N){
  vector<int> r1 = dijkstra(s,N,G), r2 = dijkstra(t,N,G);
  int l = r1[t], d1[l+1], d2[l+1];
  fill(d1,d1+l+1,0);
  fill(d2,d2+l+1,0);
  for(int i = 0; i < N; ++i){
    if(r1[i] <= l) ++d1[r1[i]];
    if(r2[i] <= l) ++d2[r2[i]];
  }
  long long int ret = 0;
  for(int i = 0; i < l-1; ++i){
    ret += (long long int)d1[i]*d2[l-2-i];
  }
  return ret;
}

int main(){
  int N, M, s, t, x, y;
  cin >> N >> M >> s >> t;
  --s;--t;
  vector< vector<pii> > G(N);
  for(int i = 0; i < M; ++i){
    cin >> x >> y;
    --x;--y;
    G[x].push_back(make_pair(1,y));
    G[y].push_back(make_pair(1,x));
  }
  cout << count(G,s,t,N) << endl;
  return 0;
}